Derive self-inductance formula from first principles.

Self-Inductance of a Solenoid — Derive L = μ₀n²Al

Question

Derive the expression for self-inductance of a solenoid having $n$ turns per unit length, cross-sectional area $A$ , and length $l$ . Show that $L = \mu_0 n^2 Al$ .

Solution — Step by Step

Let current $I$ flow through the solenoid. We need to find the flux linkage and use $L = N\Phi / I$ .

Our solenoid has $n$ turns per unit length, so total turns $N = nl$ . Cross-section area is $A$ .

The field inside a long solenoid (from Ampere’s circuital law) is uniform:

B = \mu_0 n I

This holds strictly for an ideal solenoid — infinite length, tightly wound. For our derivation, we treat it as ideal.

Magnetic flux through a single turn:

\Phi_1 = B \cdot A = \mu_0 n I A

We use $\Phi = BA\cos\theta$ with $\theta = 0°$ since $\vec{B}$ is parallel to the area vector.

Total flux linkage $\Psi$ is the sum across all $N = nl$ turns:

\Psi = N\Phi_1 = (nl)(\mu_0 n I A) = \mu_0 n^2 A l \cdot I

By definition, $L = \Psi / I$ :

L = \frac{\mu_0 n^2 A l \cdot I}{I}

\boxed{L = \mu_0 n^2 A l}

Why This Works

Self-inductance measures a coil’s “reluctance to change current” — it’s the ratio of flux linkage to current. The key insight is that $B$ inside a solenoid depends on $n$ (turns per unit length), not the total number of turns. But the flux linkage accumulates over all $nl$ turns, so $L$ ends up proportional to $n^2$ .

This $n^2$ dependence is what makes solenoids powerful inductors. Double the winding density, and $L$ goes up by four times. This is why transformer cores are wound as tightly as possible.

The formula $L = \mu_0 n^2 Al$ also tells us that $L$ is purely geometric — it depends only on the shape and winding of the solenoid, not on what current is flowing. This is a hallmark of linear inductors.

Alternative Method

You can also arrive at $L$ using the energy stored in the magnetic field.

Energy stored in volume $V = Al$ of the solenoid:

U = \frac{B^2}{2\mu_0} \cdot Al = \frac{(\mu_0 nI)^2}{2\mu_0} \cdot Al = \frac{\mu_0 n^2 I^2 Al}{2}

Since $U = \frac{1}{2}LI^2$ , comparing both sides:

L = \mu_0 n^2 Al

Same result. This energy method is faster if you already know the field, and JEE occasionally asks you to use this route instead of flux linkage — so keep it in your toolkit.

In JEE Main, if they give you total turns $N$ instead of $n$ , substitute $n = N/l$ before using the formula. Written in terms of $N$ : $L = \mu_0 N^2 A / l$ . Both forms appear in PYQs.

Common Mistake

Students write total flux as $\Phi = NBA$ and then substitute $B = \mu_0 NI$ (using total turns $N$ instead of turns per unit length $n$ ). This gives $L = \mu_0 N^2 A$ — missing the $1/l$ factor. The field inside a solenoid is $B = \mu_0 nI$ where $n = N/l$ . Mixing up $N$ and $n$ here costs you the entire derivation mark in board exams.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next